Consider the following planes. x + y + z = 5, x + 3y + 3z = 5A) Find parametric equations for the line of intersection of the planes. B) Find the angle between the planes.

Step-by-step explanation:

Consider the following planes.  x + y + z = 5, x + 3y + 3z = 5

the parametric equations for the line of intersection of the planes are determined as follows:

From the first plane, the normal of the first plane is

from the second plane, the normal of the second plane is

= i(3-3) -j(3-1)+k(3-1)

= i(0) - j(2) + k(2)

= -2j +2k

Suppose z = 0

x+y + 0 = 5  ---(1)

x+3y + 3(0) = 5 (2)

subtracting by elimination

-2y = 0

y = 0/-2

y = 0

The intersection on point of the plane is (5,0,0)

The  equation of plane is r (t) =(5, 0, 0) + t(0, -2, 2)

∴

(x(t), y (t), z(t) ) = (5, -2t, 2t)

B) Find the angle between the planes

The angle between the planes can be represented by the equation:

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !